Prove that if $\{f_k\}$ converges uniformly to $f$ and $\{g_k\}$ converges uniformly to $g$, then $\{f_k + g_k\}$ converges uniformly to $f+g$.
I came up with this proof and I was wondering if it is OK.
First of all, this is my definition of uniform convergence:
$f_n:T \rightarrow \mathbb{R}$ converges uniformly to $f:T \rightarrow \mathbb{R}$ if for each $\epsilon>0$
there exists $n(\epsilon)$ such that if $n \ge n(\epsilon)$ then $|f_n(x)-f(x)|<\epsilon$ for any $x\in T$.
We want to prove that $\{f_k+g_k\} \rightarrow f+g$ uniformly, so given $\epsilon>0$ we have to prove that $\exists n(\epsilon): \forall n\geq n(\epsilon), \ \ |(f_n(x) + g_n(x)) – (f(x)+g(x))|<\epsilon.$
Let $\epsilon>0.$ We know that $\{f_k\} \rightarrow f$ uniformly and $\{g_k\} \rightarrow g$ uniformly, so:
- $\exists n(\epsilon): \forall n\geq n(\epsilon), \ \ |f_n(x) – f(x)|
< {\epsilon \over 2}.$ - $\exists m(\epsilon): \forall m\geq m(\epsilon), \ \ |g_m(x) – g(x)|
<{\epsilon \over 2}.$
Let' s consider $t_0=max\{n, m \}$. If $t\geq t_0$, then $|(f_t(x) + g_t(x)) – (f(x)+g(x))|$ = $|(f_t(x) – f(x)) + (g_t(x)-g(x))| \leq |f_t(x) – f(x)| + |g_t(x)-g(x)| < {\epsilon \over 2}+{\epsilon \over 2}=\epsilon.$
Best Answer
Let $I $ be an intervall, $$F_n =\sup_I |f_n-f|$$ and $$G_n=\sup_I |g_n-g|.$$
then $\forall x\in I,$
$$|(f_n+g_n) (x)-(f+g)(x)|\le $$ $$|(f_n-f)(x)|+|(g_n-g)(x)|\le$$ $$ F_n+G_n $$
and
$$ H_n=\sup_I |(f_n+g_n)-(f+g)|\le$$ $$ F_n+G_n $$
but $$\lim_{n\to+\infty} F_n=\lim_{n\to+\infty} G_n=0$$ $$ \implies \lim_{n\to+\infty}H_n=0$$ by squeezing.
$\implies \; (f_n+g_n) $ converges uniformly to $f+g $ at $I $.